Which of the following numbers are divisible by 3 and by 6? (i) 197232 (ii) 321729 (iii) 197232 (iv) 1790184

To determine which numbers are divisible by both 3 and 6, we have to check each number against the divisibility rules for 3 and 6.
Divisibility Rules:

Divisibility by 3 : A number is divisible by 3 if the sum of its digits is divisible by 3.
Divisibility by 6 : A number is divisible by 6 if it is divisible by both 2 and 3 (since 6 = 2 \times 3).
For divisibility by 2, the number must be even (i.e., the last digit is 0, 2, 4, 6, or 8).



Analyzing Each Number:

197232 :

Last digit is 2 , so it is even (divisible by 2).
Sum of digits : 1 + 9 + 7 + 2 + 3 + 2 = 24
24 is divisible by 3.
Since 197232 is divisible by both 2 and 3, it is divisible by 6 .


321729 :

Last digit is 9 , so it is not even.
Since it is not divisible by 2, it cannot be divisible by 6, regardless of divisibility by 3.


197232 (same number as the first, repeating):

Analyzed above: divisible by 6 .


1790184 :

Last digit is 4 , so it is even (divisible by 2).
Sum of digits : 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30
30 is divisible by 3.
Since 1790184 is divisible by both 2 and 3, it is divisible by 6 .



Conclusion:

197232 is divisible by 3 and by 6.
321729 is not divisible by 6.
1790184 is divisible by 3 and by 6.

Thus, the correct choices are (i) 197232 , (iii) 197232 (again, due to repetition), and (iv) 1790184 .

Answered by AvaCharlotteMiller | 2025-07-21